Question: Let $y=\dfrac{1-2x}{3x^2}$. What is the value of $\dfrac{dy}{dx}$ at $x=1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac13$ (Choice B) B $0$ (Choice C) C $1$ (Choice D) D $\dfrac19$
Let's first find the expression for $\dfrac{dy}{dx}$ (i.e. for any input value $x$ ). Then, we can plug $x=1$ and evaluate. $\dfrac{1-2x}{3x^2}$ is a rational expression. To find the derivative of rational expressions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d y d x = d d x ( 1 − 2 x 3 x 2 ) = ( 3 x 2 ) d d x ( 1 − 2 x ) − ( 1 − 2 x ) d d x ( 3 x 2 ) ( 3 x 2 ) 2 = ( 3 x 2 ) ( − 2 ) − ( 1 − 2 x ) ( 6 x ) ( 3 x 2 ) 2 = − 6 x 2 − 6 x + 12 x 2 ( 3 x 2 ) 2 = 6 x 2 − 6 x ( 3 x 2 ) 2 The quotient rule Differentiate ( 1 − 2 x ) & ( 3 x 2 ) Expand \begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\dfrac{1-2x}{3x^2}\right) \\\\ &=\dfrac{(3x^2)\dfrac{d}{dx}(1-2x)-(1-2x)\dfrac{d}{dx}(3x^2)}{(3x^2)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{(3x^2)(-2)-(1-2x)(6x)}{(3x^2)^2}&&\gray{\text{Differentiate }(1-2x)\text{ & }(3x^2)} \\\\ &=\dfrac{-6x^2-6x+12x^2}{(3x^2)^2}&&\gray{\text{Expand}} \\\\ &=\dfrac{6x^2-6x}{(3x^2)^2} \end{aligned} So we found that $\dfrac{dy}{dx}=\dfrac{6x^2-6x}{(3x^2)^2}$. Now let's plug $x= 1$ : $\begin{aligned} &\phantom{=}\dfrac{6( 1)^2-6( 1)}{(3( 1)^2)^2} \\\\ &=\dfrac{6-6}{(3)^2} \\\\ &=0 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $x=1$ is $0$.